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convert 220 AC to 3.6 VDC for lighting LED

Discussion in 'General Electronics Chat' started by mamun2a, Feb 24, 2008.

  1. mamun2a

    mamun2a New Member

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    i will be grateful if any one can provid a diagram for using LED in 220 VAC with only Resistor and diode.

    thanks to all
    mamun2a at gmail.com, dhaka
     
  2. Hero999

    Hero999 Banned

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    No diagram is required.

    Connect the diode in reverse parallel with the LED and connect a huge 5W 10k resistor in series.
     
  3. Diver300

    Diver300 Well-Known Member

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    If you use a 0.22 :mu:F capacitor (non polarised and rated to at least 300 V) then the current will be the same and there won't be nearly as much power dissapated.
     
  4. dave

    Dave New Member

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  5. mneary

    mneary New Member

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    The capacitor will be seeing up to 339 volts in normal operation, so a 300V device is clearly inadequate. The capacitor rating needs to be rated at least 600V, and should also be rated for connection to the AC mains.

    You still need a small series resistor to limit the current when there are glitches on the line.
     
  6. on1aag

    on1aag New Member

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    Hi Diver300,

    You need to use a "X2" rated capacitor, a non polarized capacitor rated
    for 300 volt is not designed for 230 Vac/50Hz operation, check the datasheet
    for this capacitor. You'll also need a bleeder resistor to discharge the
    capacitor when power is switched off, preferably a high voltage resistor
    (about 10 Meg is ok).
    And you will also need to connect a resistor in series with this capacitor
    to limit the inrush current to a safe value for the led/diode, preferably
    a fusible resistor (select a value between 1k and 2k2/0.5W). These "special"
    resistors cost a bit more than the regular resistors but if you want to be
    safe at all times it is money well spent.

    on1aag.
     
  7. MrNobody

    MrNobody New Member

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    What...???
    You can light up a 3.6V DC LED directly from 220VAC just by using some diodes, capacitor and resistor without using any sort of step down transformer..?
    WOW...
    Sorry.. I am new to this.. umm.. can somebody kindly explain the concept or point me to where I can know more about this..?
    Thanks..
     
  8. on1aag

    on1aag New Member

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    Hi MrNobody,

    Try this, but don't try this at your home. :D

    on1aag.
     

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  9. audioguru

    audioguru Well-Known Member Most Helpful Member

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    The LED is powered with half-wave so it will flicker at 25Hz and drive you crazy.

    Use a rectifier bridge to eliminate the flickering and then a filter capacitor can also be used. The series capacitor that limits the current with its capacitive reactance will then need to be half the value before.
     
  10. mneary

    mneary New Member

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    It flickers at 50 Hz, but will still drive you (me) crazy. Full wave flickers at 100 Hz, which tends not to bother me.
     
  11. audioguru

    audioguru Well-Known Member Most Helpful Member

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    You are absolutely correct and I was a dumbo.
    Thanks for the correction.
     
  12. on1aag

    on1aag New Member

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    Alright, alright, have it your way.
    Why do I have to do all the work while you sit back and relax ?
    Is it because I'm stupid ?

    on1aag.
     

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  13. audioguru

    audioguru Well-Known Member Most Helpful Member

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    I wish my LED Christmas tree lights are smooth like that.
     
  14. MrNobody

    MrNobody New Member

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    Interesting..
    I see.. most of the voltage drop is accross the 10M resistor.. The LED only receives a small portion of the 220V voltage..

    Hmm.. jst wondering.. how about powering the PIC MCU using this method..?
    Attaching a full bridge rectifier circuit between the 10M resistor and 1K resistor, 5V voltage regulator and capacitors to smooth the ripples..

    What are the disadvantages of that apart from accidentally touching the 10M resistor and high power dissipation..?

    Just curious..
     
  15. Hero999

    Hero999 Banned

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    C1 needs to be rated for at least 25V.

    The power dissipation is actually pretty low, about 300mW.

    The apparant power is 3.3VA.

    The power factor is very poor at only 0.091. You could improve this but you'd need to add a large inductor which would be very bulky and defeat the purpose of having a light weight power supply.
     
  16. DMW

    DMW New Member

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    Actually that's not true I believe [for the first diagram]

    The 10M has little effect on the circuit, you see capacitors have a resistance to low frequencies, this is called reluctance rather than resistance because unlike resistance no power is dissipated in the capacitor.

    The Formula shown below determines the reluctance for a capacitance.
    Xc = 1/(2 * Pi * F * C)
    Xc = 1/(2*Pi*50*220nF)
    Xc = 14.4Kohm

    This is in parallel with 10M, because 10M is so high it makes little difference to the resistance.

    14.4K + 1K because its in series with the 1K resistor is
    15.4K
    current = 230 / 15.4 = 15mA

    Seems a little low to me.

    Edit:Whops I was using RMS value
    root(2) * 230 = 425V
    425 / 15400 = 21mA.

    Thats better :)
     
    Last edited: Feb 29, 2008
    • Like Like x 1
  17. ecerfoglio

    ecerfoglio New Member

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    It's not reluctance but reactance.

    Both capacitors (Xc = 1 / (2 pi F C) and inductors (XL = 2 pi F L) have reactance. If you have an inductor and a capacitor in series the total reactance is de difference of them (Total X = XL - Xc)

    (The Reluctance is used in magnetic circuits)
     
  18. DMW

    DMW New Member

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    ops yes i did know its reactants, I must of chosen the wrong one without thinking when dong spell check. :)

    edit

    I thought the total reactants of L/C was:
    square-root(Xc^2 + XL^2)

    Like Pythagoras theorem

    Or is that something else?
     
    Last edited: Feb 29, 2008
  19. ecerfoglio

    ecerfoglio New Member

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    Not, if you have a capacitor, an inductor and a resistor in series, then the impedance (Z) is

    Z = square-root(R^2 + (XL - Xc)^2)

    (or, using complex numbers, Z = R + j (XL - XC), where j is the square root of minus one)
     
  20. DMW

    DMW New Member

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    Of course its in converting from rectangular to polar form.
     
  21. RODALCO

    RODALCO Well-Known Member

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    230 Volts LED

    I have posted a schema a while back under 230 Volts LED with photo's.

    I have used this succesfully for nearly 20 years in industrial metering and indication circuits on power systems 230 / 240 Volts 50 Hz.

    D1 - 1N4007
    D2 - 1N4148, 1N914
    D3 - High efficiency LED
    R1 - R2 - 33 or 39 k:eek:hm: 1 Watt resistor

    D1 assisits in reducing the power dissipation in R1 and R2 to 0.7.

    To reduce flicker if it bothers you use a bridge rectifier as already suggested.
     

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